## Chemistry: The Central Science (13th Edition)

The pH of this solution is equal to $4.87$.
1. Find the numbers of moles: - $50ml = 0.05L$ $C(CH_3COOH) * V(CH_3COOH) = 0.15* 0.05 = 7.5 \times 10^{-3}$ moles $C(CH_3COO^-) * V(CH_3COO^-) = 0.2* 0.05 = 0.01$ moles 2. Calculate the total volume: - Total volume: 0.05 + 0.05 = 0.1L 3. Calculate the final concentrations: $[CH_3COOH] : \frac{7.5 \times 10^{-3}}{0.1} = 0.075M$ $[CH_3COOH] : \frac{0.01}{0.1} = 0.1M$ 4. Calculate the pKa Value $pKa = -log(Ka)$ $pKa = -log( 1.8 \times 10^{- 5})$ $pKa = 4.745$ 5. Check if the ratio is between 0.1 and 10: - $\frac{[Base]}{[Acid]} = \frac{0.1}{0.075}$ - 1.333: It is. 6. Check if the compounds exceed the $K_a$ by 100 times or more: - $\frac{0.1}{1.8 \times 10^{-5}} = 5556$ - $\frac{0.075}{1.8 \times 10^{-5}} = 4167$ 7. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 4.745 + log(\frac{0.1}{0.075})$ $pH = 4.745 + 0.1249$ $pH = 4.87$